What about regular rings with finite Krull dimension? Not all such rings are contained in your list above.
–
Donu ArapuraJan 24 '11 at 22:25

In case you aren't sure about how to get such examples as mentioned in my last comment, take the coordinate ring of any nonsingular affine variety.
–
Donu ArapuraJan 24 '11 at 22:40

That's the answer I was looking for. (I was only familiar with the noncommutative case, and didn't know the term "regular ring" or Serre's characterization until about ten seconds ago.) So they're actually incredibly common. Sorry, I didn't realize the question was so easy or well known.
–
arsmathJan 24 '11 at 22:40

Donu if you put your answer as an answer, I'll accept it so that this question doesn't linger on the unanswered question list.
–
arsmathJan 24 '11 at 22:42

2 Answers
2

At arsmath's request, I'm making this official.
(This is pretty standard commutative algebra, but I realize not everyone has gone through it.)

A commutative ring $R$ is regular if it's noetherian and its local rings are regular.
Using Serre's theorem e.g. Matsumura Commutative Ring Theory p 156, and the fact
that $Ext$ commutes with localization, we can see that any regular ring with finite Krull
dimension has finite global dimension.

To an algebraic geometer regular = nonsingular. So in particular, so there is a large
supply of basic examples arising as coordinate rings of nonsingular affine varieties.
This is a bit circular the way I'm saying it, but of course, you can test the condition
using the Jacobian criterion...

An artinian ring $R$ is a finite direct product of local artinian rings. If $R$ is of finite global dimension, so are the factors, and then they are regular local by Serre's theorem. As regular local rings are domains, the Jacobson radical of the factors has to be trivial (for its elements are nilpotent) and then $R$ is a product of fields.